document.write( "Question 252022: (2-4i)/(2)
\n" ); document.write( "(2+4i)/(-5+2i)
\n" ); document.write( "(1-i)/(-4-6i)
\n" ); document.write( "(3+5i)/(-2+i) \n" ); document.write( "
Algebra.Com's Answer #183859 by jsmallt9(3758)\"\" \"About 
You can put this solution on YOUR website!
The standard form for complex numbers is a + bi. So in each problem we want to end up with an expression of this form.

\n" ); document.write( "\"%282-4i%29%2F2\"
\n" ); document.write( "All we have to do here is split the fraction into two parts:
\n" ); document.write( "\"2%2F2+-+4i%2F2+\"
\n" ); document.write( "and simplify:
\n" ); document.write( "\"1+-+2i\"
\n" ); document.write( "and rewrite as an addition:
\n" ); document.write( "\"1+%2B+%28-2i%29\"

\n" ); document.write( "\"%282%2B4i%29%2F%28-5%2B2i%29\"
\n" ); document.write( "\"i+=+sqrt%28-1%29\". So we can look at this problem as either division of complex numbers or as rationalizing the denominator of a fraction. Either way we do the same thing. Using the pattern \"%28a%2Bb%29%28a-b%29+=+a%5E2+-+b%5E2\" we will turn the binomial (2-term expression) denominator into a binomial of perfect squares. Expressions of the form a+b and a-b are called conjugates. And this pattern tells us that if you multiply conjugates that you get the difference (subtraction) of the square of the first term and the square of the second term. This is a very handy way to eliminate square roots in binomials.

\n" ); document.write( "So we will multiply the numerator and denominator of
\n" ); document.write( "\"%282%2B4i%29%2F%28-5%2B2i%29\"
\n" ); document.write( "by the conjugate of the denominator: -5-2i
\n" ); document.write( "\"%28%282%2B4i%29%2F%28-5%2B2i%29%29%28%28-5-2i%29%2F%28-5-2i%29%29\"
\n" ); document.write( "Using FOIL on the top and FOIL or the pattern on the bottom we get:
\n" ); document.write( "\"%28-10+-+4i+-+20i+-8i%5E2%29%2F%28%28-5%29%5E2+-+%282i%29%5E2%29\"
\n" ); document.write( "which simplifies to:
\n" ); document.write( "\"%28-10+-+4i+-+20i+-8i%5E2%29%2F%2825+-+4i%5E2%29\"
\n" ); document.write( "Since \"i%5E2+=+-1\":
\n" ); document.write( "\"%28-10+-+4i+-+20i+-8%28-1%29%29%2F%2825+-+4%28-1%29%29\"
\n" ); document.write( "\"%28-10+-+4i+-+20i+%2B8%29%2F%2825+%2B+4%29\"
\n" ); document.write( "Combining like terms:
\n" ); document.write( "\"%28-2+-+24i%29%2F29\"
\n" ); document.write( "Now we can split this into two terms:
\n" ); document.write( "\"-2%2F29+-+24i%2F29\"
\n" ); document.write( "In a + bi form:
\n" ); document.write( "\"-2%2F29+%2B+%28%28-24%29%2F29%29i\"

\n" ); document.write( "The last two problems
\n" ); document.write( "(1-i)/(-4-6i)
\n" ); document.write( "(3+5i)/(-2+i)
\n" ); document.write( "are just like this one. I'll leave them up to you. Just multiply the numerator and denominator by the conjugate of the denominator.

\n" ); document.write( "P.S. Using conjugates like this is also useful for rationalizing denominators like:
\n" ); document.write( "\"%282+%2B+sqrt%283%29%29%2F%285+-+sqrt%282%29%29\"
\n" ); document.write( "
\n" ); document.write( "
\n" ); document.write( "\"%2810+%2B+2sqrt%282%29+%2B+5sqrt%283%29+%2B+sqrt%286%29%29%2F%2825+-+2%29\"
\n" ); document.write( "\"%2810+%2B+2sqrt%282%29+%2B+5sqrt%283%29+%2B+sqrt%286%29%29%2F23\"
\n" ); document.write( "which may look like a mess but at least it's a mess with a rational denominator.\r
\n" ); document.write( "\n" ); document.write( " \n" ); document.write( "
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